Question

2)

a) Calculate the change in pH when 8.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

delta pH= ?

b) Calculate the change in pH when 8.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

delta pH= ?

Answer 1

(a) Find out pH of buffer solution using Hederson - Hasselbalch equation-

pOH = pK_b   + log { [salt] / [ Base ] }

K_b   of NH₄  OH at room temperature = 1.80 x 10^-5 ( Refer table for dissociation constant of bases )

      hence , pK_{b =} - log K_b   = - log (1.80 x 10^-5 ) = 4.7447

Now, p (OH) = 4.7447 + log ( 0.1 / 0.1 )

= 4.7447

& pH = ( 14.00 - 4.7447) = 9.2553

Now, when 8.00 ml . of 0.100 M HCl solution is added, the total volume of solution becomes 108.00ml and the concentration of acid would be

108 x [ new conc. of acid say ,c ] = 100 x0.1

c = (100/108) (0.1 )

=0. 09259 M

hence decrease in conc. of HCl = (0.10 - .092590) = 0.0074 moles / l

HCl ionises completely to form Cl -, hence this would also cause an increase in conc. of salt = ( 0.10 + 0.0074) = 0.10074 moles /l

Apply Handerson's equation again to get the changed or new p(OH)

p(OH ) = 1.8 x 10 ^-5   + log 0.09259 / 0.10074)

= 5.714

hence pH = 14- 5.714 = 8.286

delta pH = 9.2553 - 8.286 = )>9693

or = 0.97

   Similarly the other part of the question should be attempted